

A068069


a(n) = least k which is the start of n consecutive integers each with a different number, 1 through n, of distinct prime factors.


3




OFFSET

0,2


COMMENTS

a(n) >= n!. If the canonical factorization of k is the product of p^e(p) over primes, then the number of distinct number of prime factors is simply the count of the number of p's.


LINKS

Table of n, a(n) for n=0..9.
P. Erdős, Megjegyzések a Matematikai Lapok két problémájához (Remarks on two problems, in Hungarian), Mat. Lapok 11 (1960), pp. 2632.
J.M. De Koninck, J. B. Friedlander, and F. Luca, On strings of consecutive integers with a distinct number of prime factors, Proc. Amer. Math. Soc., 137 (2009), 15851592.


FORMULA

Koninck, Friedlander, & Luca prove that a(n) > exp(2n + o(n)), but note that an earlier result of Erdős is "essentially equivalent".  Charles R Greathouse IV, Feb 04 2013


EXAMPLE

a(1) = 2 because 2 has the single prime factor 2; a(2) = 5 because 5 = 5^1 & 6 = 2*3 which have 1 & 2 prime factors respectively; a(3) = 28 because 28 = 2^2*7^1, 29 = 29^1 & 30 = 2*3*5 which have 2, 1 & 3 prime factors respectively; a(4) = 417 because 417 = 3*139, 418 = 2*11*19, 419 = 419^1 & 420 = 2^2*3*5*7 which have 2, 3, 1 & 4 prime factors (distinct) respectively and this represents a recordbreaking number.


MATHEMATICA

k = 3; Do[k = k  n; a = Table[ Length[ FactorInteger[i]], {i, k, k + n  1}]; b = Table[i, {i, 1, n}]; While[ Sort[a] != b, k++; a = Drop[a, 1]; a = Append[a, Length[ FactorInteger[k]]]]; Print[k  n + 1], {n, 1, 7}]


CROSSREFS

Cf. A067665.
Sequence in context: A324264 A138293 A316972 * A292499 A306893 A105787
Adjacent sequences: A068066 A068067 A068068 * A068070 A068071 A068072


KEYWORD

more,nonn


AUTHOR

Robert G. Wilson v, Feb 20 2002


EXTENSIONS

One more term from Labos Elemer, May 26 2003
One more term from Donovan Johnson, Apr 03 2008
Corrected example and a(9) from Donovan Johnson, Aug 31 2010


STATUS

approved



